Question

因式分解解一元二次方程1.(2x-1)²=(5-x)²2.x²-4x+4=(5-2x)&s？

Answer 1

1.(2x-1)²=(5-x)²
(2x-1)^2-(5-x)^2=0,
(2x-1+5-x)(2x-1-5+x)=0,
(x+4)(3x-6)=0
x+4=0,3x-6=0,
x1=-4,x2=2,
2.x²-4x+4=(5-2x)²,
配方法
x^2-4x+4=25-20x+4x^2
x^2-4x^2-4x+20x+4-25=0,
-3x^2+16x-21=0,
3x^2-16x+21=0,
x^2-16x/3+7=0,
x^2-16x/3=-7,
x^2-16x/3+64/9=-7+64/9
(x-8/3)^2=1/9,
x-8/3=1/3,x-8/3=-1/3
x1=3,x2=7/3
公式法
x^2-4x+4=25-20x+4x^2
x^2-4x^2-4x+20x+4-25=0,
-3x^2+16x-21=0,
3x^2-16x+21=0,
a=3,b=-16,c=21,
b^2-4ac
=256-252
=4,
x=（16±2）/6
x1=3,x2=7/3
3.(2x+1)(2x+3)=15
4x^2+8x+3=15,
4x^2+8x+3-15=0,
4x^2+8x-12=0,
x^2+2x-3=0
(x+3)(x-1)=0,
x+3=0,x-1=0,
x1=-3,x2=1
4.(x+5)²-2(x+5)-8=0
(x+5-4)(x+5-2)=0,
(x+1)(x+3)=0,
x+1=0,x+3=0,
x=-1,x=-3,1,因式分解解一元二次方程
1.(2x-1)²=(5-x)²
2.x²-4x+4=(5-2x)²(同时用配方法、公式法解出)
3.(2x+1)(2x+3)=15
4.(x+5)²-2(x+5)-8=0
非常感谢各位!